Question

How do I prove the relation between $AM,GM$ and $HM?$

Answer 1

Step 1. The relation between $AMGMHM$ can be represented by the formula $G{M}^2=AM\times HM$.

Here the product of the arithmetic mean$\left(AM\right)$ and harmonic mean$\left(HM\right)$ is equal to the square of the geometric mean$\left(GM\right)$

Now, consider RHS$=AM\times HM$ ….(1)

Step 2. Let us consider $a,AM,b$ is an Arithmetic progression.

Common difference, $d=AM-a$

$=b-AM$

Or

$2AM=a+b$

$\Rightarrow$$AM=\frac{a+b}{2}$

Step 3. let the reciprocal of Harmonic progression $a,HM,b$ is the Arithmetic progression.

$\frac{1}{a},\frac{1}{HM},\frac{1}{b}$ is an AP

For above AP, the common difference, $d=\frac{1}{HM}-\frac{1}{a}$

$=\frac{1}{b}-\frac{1}{HM}$

⇒ $\frac{2}{HM}=\frac{1}{b}+\frac{1}{a}$

⇒ $\frac{2}{HM}=\frac{\left(a+b\right)}{ab}$

$\Rightarrow$ $\frac{1}{HM}=\frac{\left(a+b\right)}{2ab}$

$\Rightarrow$ $HM=\frac{2db}{\left(a+b\right)}$

Step 4. Put the values of $\left(AM\right)$ and $\left(HM\right)$ in equation (1), we get

$\Rightarrow$ RHS$=AM\times HM$

$=\frac{\left(a+b\right)}{2}\times \frac{2ab}{\left(a+b\right)}$

$=\frac{\left(a+b\right)2ab}{2\left(a+b\right)}$

$\Rightarrow$ RHS$=ab$

Now, Consider LHS$=G{M}^2$ …(2)

Step 5. Let $a,GM,b$ is the geometric progression.

Common ratio, $\mathit{r}\mathbf{=}\frac{\mathbf{G}\mathbf{M}}{\mathbf{a}}$

$=\frac{b}{GM}$

$\Rightarrow$ $\frac{GM}{a}=\frac{b}{GM}$

$ab=G{M}^2$

$\Rightarrow$ LHS$=ab$

$\Rightarrow$ LHS$=$RHS

From equation (2) and (3), we get

$\mathit{G}{\mathit{M}}^{\mathbf{2}}\mathbf{=}\mathit{A}\mathit{M}\mathbf{}\mathbf{\times }\mathbf{}\mathit{H}\mathit{M}$

Hence, Proved.